Answers
Let us assume to the contrary that √3 is a rational number.
It can be expressed in the form of p/q
where p and q are co-primes and q≠ 0.
⇒ √3 = p/q
⇒ 3 = p²/q² (Squaring on both the sides)
⇒ 3q² = p²………………………………..(1)
It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.
So we have p = 3r
where r is some integer.
⇒ p² = 9r²………………………………..(2)
from equation (1) and (2)
⇒ 3q² = 9r²
⇒ q² = 3r²
We have two cases to consider now.
Case I
Suppose that r is even. Then r² is even, and 3r² is even which implies that q² is even and so q is even, but this cannot happen. If both q and r are even then gcd(q,r)≥² which is a contradiction.
Case II
Now suppose that r is odd. Then r² is odd and 3r² is odd which implies that q² is odd and so q is odd. Since both q and r are odd, we can write q=2m−1 and r=2n−1 for some m,n∈N.
Therefore
q²=3r²
(2m−1)²=3(2n−1)²
4m²−4m+1=3(4n²−4n+1)
4m²−4m+1=12n²−12n+3
4m²−4m=12n²−12n+2
2m²−2m=6n²−6n+1
2(m²−m)=2(3n²−3n)+1
We note that the lefthand side of this equation is even, while the righthand side of this equation is odd, which is a contradiction. Therefore there exists no rational number r such that r2=3.
Hence the root of 3 is an irrational number.
Hence Proved✅